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I am using a voltage divider to supply a voltage to a system (I don't know what sort of circuit I am supplying the signal to, except it's used to drive a very large motor. It needs a 0-5v signal). Other than this, there is another input to the system as well, a 72V switching signal.

I am using two 10K resistors at the moment (this is for testing purposes) for the voltage divider. The voltage divider has a supply of 5V, so the signal is 2.5V. The circuit is given below, and the red arrow indicates the signal.

voltage divider circuit

The strange behavior occurs when the switching signal is turned on, at which point the voltage signal suddenly jumps to 5V.

In addition to that, there is no response to the input signal at all. The system is working, because it responds to the original signal from an accelerator pedal.

Any idea what could be causing this behavior?

Thank you.

EDIT: The original accelerator pedal is the one in this link: https://www.alibaba.com/product-detail/electric-vehicle-foot-pedal-with-potentiometer_60649371220.html?spm=a2700.7724857.main07.9.28a05756UO1eV2.

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    \$\begingroup\$ Any load put on that will affect the voltage. It effectively puts R2 in parallel so you will have to account for that. Or put a buffer in \$\endgroup\$
    – MCG
    May 24, 2018 at 7:54
  • \$\begingroup\$ if 'It needs a 0-5v signal', why are you giving it 2.5v? As that 2.5v is coming from an impedance of 5k, is the unknown system you're driving high enough impedance to accept that, or is it dragging it somewhere with a lower load? What's the circuit of your original accelerator pedal that does work? \$\endgroup\$
    – Neil_UK
    May 24, 2018 at 7:55
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    \$\begingroup\$ Any idea what could be causing this behavior? No because you show only a very limited part of your setup. I can think of 100 circuits I can connect to that 2.5 V output which will all cause it to become 5 V. Question is, what is connected in your setup? The "strangeness" is not in the circuit, it is in your perception of what you think the circuit should do and what it actually does (which depends a lot on what is connected to it). \$\endgroup\$ May 24, 2018 at 7:55
  • \$\begingroup\$ I couldn't find any information on the circuit of the original accelerator pedal. It's the pedal in this link: alibaba.com/product-detail/…. \$\endgroup\$
    – 884-15-X-C
    May 24, 2018 at 8:18
  • \$\begingroup\$ I have also tried driving it using the output from an LM324N op amp (no response) and with a power transistor (again, no response). With these inputs, the voltage doesn't go to 5V when the switching signal is on. \$\endgroup\$
    – 884-15-X-C
    May 24, 2018 at 8:22

4 Answers 4

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Without knowing your load circuit, I can still say it's most likely an impedance problem. You have two options. Either you can decrease the output impedance, or increase the load impedance. We can't help you with the latter unless you give us more information.

To decrease the output impedance, you could replace your voltage divider with a linear regulator like the CUI P7805-Q12-S2-S, or keep the voltage divider and add an op-amp configured for non-inverting buffer mode.

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  • \$\begingroup\$ I tried using an op-amp in non-inverting buffer mode with an analog signal, but there is no response from the circuit. It doesn't jump to 5V with the op-amp buffer, though. \$\endgroup\$
    – 884-15-X-C
    May 24, 2018 at 8:32
  • \$\begingroup\$ Then it sounds like you've potentially solved one problem and you're on to the next one 😀 \$\endgroup\$
    – Reinderien
    May 24, 2018 at 8:33
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You can't do that: if you have any load connected to your voltage divider (and a large motor as an particularly heavy load), you don't get a voltage divider, you get a loaded voltage divider, as you must imagine the load to be in parallel with either of your resistors. Linear networks basics!

What you need is a voltage regulator of kinds, not a voltage divider.

For heavy loads, you probably don't want to use a linear voltage regulator, though your voltages really indicate you don't have much of a choice?

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  • \$\begingroup\$ From what I can read in the question, this signal is unlikely to drive the motor, considering it has a separate 72 volt input. \$\endgroup\$
    – pipe
    May 24, 2018 at 8:31
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Connecting ground references.

There is not enough context and no schematic in your question so we can only guess at possible problems. Your diagram does not show that you have referenced your divider's ground to the controller ground. If you have failed to do this then your control voltage has no return path and will not work.

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I suspect you are connecting it incorrectly. Normally the control input to a moter speed control will look something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

If you connect the voltage divider to A rather than B, it will behave as you suggest. The value of R1 is not necessarily 100K, it could be something else, and it will load your voltage divider, but it should not load a 5K source resistance (as you have) by a huge amount.

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